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Dust: conservation of rho n tensor n is equivalent to continuity and inertial flow

Statement

Let Ω⊆A be open, let ρ:Ω→R be C1 with ρ>0, let n:Ω→V be a C1 field which is future timelike and satisfies η(n,n)=−1 everywhere (a future unit timelike field, Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(c)), and set Tab:=ρ nanb (a C1 symmetric (2,0) field). Then, on Ω, (a) ∂aTab=0 if and only if both ∂a(ρna)=0 (continuity) and na∂anb=0 (inertial flow); (b) if ρ≥0 is allowed to vanish, the implication ∂aTab=0⇒na∂anb=0 is asserted only on the open set {ρ>0}, and no assertion is made where ρ=0; (c) Under the conservation hypothesis ∂aTab=0 in (a), if J is an interval and x:J→Ω is a C2 curve with x′(s)=n(x(s)) for all s (an integral curve of n), then x′′(s)=0: the flow lines of an inertial unit field are affine straight lines in the affine structure of A. For the nonnegative-density branch in (b), the same conclusion holds for an integral curve whose image lies in {ρ>0}, under conservation there. The lemma is the exact mathematical content of the dust equations; it asserts no matter model, no interaction and no empirical fact. No choice principle is used.

Facts & Assumptions

Given: The affine Minkowski model (A,V,η,C+) with fixed inertial chart x and coordinate partial derivatives ∂a (Oriented time-oriented affine Minkowski space (the spacetime primitives), Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Directional derivatives and partial derivatives of a map U⊆Rm→Rn), an open Ω⊆A, C1 fields ρ:Ω→R and n:Ω→V with η(n,n)=−1; na:=ηabnb and indices are raised and lowered with the constant η (Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(c)). For a C1 field n the expression na∂anb is the coordinate form of the flat covariant derivative ∇nn (The flat covariant derivative on affine Minkowski space and fields along curves). For a C1 symmetric type-(2,0) field T one writes fb=∂aTab and calls T conserved when f=0 (C^k tensor fields on affine Minkowski space and their inertial components).

[F2]

If I⊆R is an interval and g:I→R is continuous on I and differentiable at every interior point with g′=0 there, then g is constant on I. (A function continuous on an interval I whose derivative vanishes at every interior point of I is constant on I; consequently two such functions with the same derivative differ by a constant)

[F3]

Chain rule: a C1 curve composed with a C1 field is C1 and its derivative is obtained by composing the derivative of the field with the curve derivative; concretely ddsnb(x(s))=∂anb(x(s)) x′a(s). (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a))

Proof

technique · direct
1.1F1given

Algebraic identity. Since ρ and n are C1, [F1] gives ∂aTab=∂a(ρnanb)=C nb+ρ na∂anb with C:=∂a(ρna), the continuity expression.

1.2given

Contraction identity. Because nbnb=η(n,n)=−1 is constant, its derivative vanishes: 0=∂a(nbnb)=2nb∂anb, hence multiplying by na and summing, nanb∂anb=0; likewise nbC nb=−C and nb na∂anb=0.

2.1step 1.1step 1.2

Forward direction of (a) and clause (b). Assume ∂aTab=0 on Ω. Contracting the identity of step 1.1 with nb and using step 1.2 gives 0=C nbnb+ρ nbna∂anb=−C, hence C=0. Substituting back gives ρ na∂anb=0 for every b. If ρ>0 everywhere, division by ρ gives na∂anb=0 on Ω; if only ρ≥0 is assumed, the same conclusion holds on the open set {ρ>0} and no claim is made where ρ=0.

2.2step 1.1

Reverse direction of (a). Conversely, if C=∂a(ρna)=0 and na∂anb=0 on Ω, then step 1.1 gives ∂aTab=0 on Ω.

3.1F2F3step 2.1

Clause (c). Let x:J→Ω be a C2 curve with x′(s)=n(x(s)). By [F3], x′′b(s)=∂anb(x(s)) x′a(s)=na∂anb(x(s)), which vanishes under the conservation hypothesis by step 2.1 (or, in the nonnegative-density branch, at every s with x(s)∈{ρ>0} by step 2.1 applied there). Hence every component (xb)′ has zero derivative at every interior point of J, and [F2] gives that (xb)′ is constant; applying [F2] to the function s↦xb(s)−xb(s0)−(xb)′(s0)(s−s0), whose derivative vanishes, gives xb(s)=xb(s0)+(xb)′(s0)(s−s0) for all s∈J and every b. Thus x is an affine straight line and x′′=0 on J.

4.1step 2.1step 2.2step 3.1∎

Steps 2.1 and 2.2 prove (a), step 2.1 proves the restricted implication (b), and step 3.1 proves (c) and its branch on {ρ>0}. The argument is a pointwise component computation with the product rule and the constancy of η(n,n); it asserts no matter model, no interaction, no dynamics of ρ beyond the displayed equations and no empirical fact, and it uses no choice principle.

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